
Energy can take many different forms: mechanical, radiant, nuclear, electrical, chemical, and thermal. Mechanical energy includes kinetic energy and potential energy (both gravitational and elastic). Work and energy are closely related, because energy is the ability to do work, and doing work on an object changes its energy. Potential energy is usually calculated in terms of a reference frame, in which a position of zero energy must be defined. Energy and work are both measured in joules (J).

mechanical energy, work, joule (J), kinetic energy, potential energy, gravitational potential energy, reference frame, elastic potential energy, spring constant


 $${E}_{k}=\frac{1}{2}m{v}^{2}$$


 $${E}_{p}=\frac{1}{2}k{x}^{2}$$



Review problems and questions 

 What forms of energy are found in
 the Sun?
 a car’s engine?

 Nuclear energy in the core, thermal energy throughout, radiant energy expanding outward from the core (as well as the surface), and mechanical energy in the internal motions of the gas are found in the Sun.
 Chemical energy from the fuel, mechanical energy from the motion of the pistons, electrical energy from the battery (as well as the alternator), thermodynamic energy that pushes the pistons, and thermal energy given off as heat are found in a car’s engine.

 A 30 kg boy sitting on a 10 kg gokart wants to travel down a hill of height 10 m, starting from rest.
 What is the initial potential energy of the boy and gokart together?
 What is their initial kinetic energy?

Answer:  3,920 J
 0
Solution: Asked: initial potential energy E_{p,i} of the boy and the gokart
Given: boy and cart combined mass m = 40 kg; hill height h = 10 m; acceleration due to gravity g = 9.8 m/s^{2} Relationships: E_{p} = mgh Solve: $$\begin{array}{lll}{E}_{p}\hfill & =\hfill & mgh\hfill \\ \hfill & =\hfill & \left(40\text{\hspace{0.17em}}\text{kg}\right)\times \left(9.8\text{\hspace{0.17em}}\text{m}/{\text{s}}^{\text{2}}\right)\times \left(10\text{\hspace{0.17em}}\text{m}\right)=3,920\text{\hspace{0.17em}}\text{J}\hfill \end{array}$$Answer: Their initial potential energy is 3,920 J.  Their initial velocity is 0, so their initial kinetic energy is also 0.

 A 1,000 kg car traveling 15.0 m/s brakes and comes to a stop after traveling 20.0 m.
 What is the car’s initial kinetic energy?
 What is the car’s final kinetic energy?
 How much work does it take to stop the car?
 How much constant force is applied in bringing the car to a stop?

Answer:  113 kJ
 0
 113 kJ
 5,630 N
Solution: Asked: initial kinetic energy E_{k,i} of the car
Given: car mass m = 1,000 kg; car initial velocity v_{i} = 15 m/s Relationships: E_{k} = ½mv^{2} Solve: $$\begin{array}{lll}{E}_{k}\hfill & =\hfill & \frac{1}{2}m{v}^{2}\hfill \\ \hfill & =\hfill & \frac{1}{2}\left(1,000\text{\hspace{0.17em}}\text{kg}\right)\times {\left(15\text{\hspace{0.17em}}\text{m}/\text{s}\right)}^{2}=112,500\text{\hspace{0.17em}}\text{J}\hfill \end{array}$$Answer: The car’s initial kinetic energy is 112,500 J.  At the end, the car is not moving, so its kinetic energy is zero.
 The brakes did enough work to reduce the car’s kinetic energy to zero. Therefore, they did 112,500 J of work.
 Asked: force F applied by the brakes
Given: work done W = 112,500 J; work done over distance d = 20 m Relationships: W = Fd Solve: Rearrange the equation W = Fd to solve for F:$$\begin{array}{lll}F\hfill & =\hfill & \frac{W}{d}\hfill \\ \hfill & =\hfill & \frac{112,500\text{\hspace{0.17em}}\text{J}}{20\text{\hspace{0.17em}}\text{m}}=5,625\text{\hspace{0.17em}}\text{N}\hfill \end{array}$$Answer: The brakes applied 5,625 N of force.

 A baseball outfielder catches a fly ball traveling 25.0 m/s with his gloved hand. When he catches the 0.140 kg baseball, the outfielder’s hand recoils by 10.0 cm. How much force did the outfielder’s hand exert in catching the ball?

Answer: The outfielder exerts 438 N of force on the ball.
Asked: Force F done by the hand on the ball Given: ball mass m = 0.140 kg; ball velocity v = 25.0 m/s; distance d = 10.0 cm = 0.100 m moved by player’s hand Relationships: E_{k} = ½mv^{2}, W = Fd Solve: Calculate the kinetic energy of the ball before being caught:$$\begin{array}{l}{E}_{k}=\frac{1}{2}m{v}^{2}=\frac{1}{2}\left(0.140\text{\hspace{0.17em}}\text{kg}\right)\times {\left(25.0\text{\hspace{0.17em}}\text{m}/\text{s}\right)}^{2}=43.75\text{\hspace{0.17em}}\text{J}\\ W=Fd\end{array}$$Rearrange the equation W = Fd to solve for F:$$\begin{array}{lll}F\hfill & =\hfill & \frac{W}{d}\hfill \\ \hfill & =\hfill & \frac{43.75\text{\hspace{0.17em}}\text{J}}{0.100\text{\hspace{0.17em}}\text{m}}=437.5\text{\hspace{0.17em}}\text{N}\hfill \end{array}$$

 A horizontal spring with k = 100 N/m is compressed by 10 cm by a 100 g mass.
 How much elastic potential energy does the compressed spring store?
 How much elastic potential energy would the compressed spring store if it were compressed the same distance by a 300 g mass?

Answer:  0.5 J
 0.5 J
Solution: Asked: elastic potential energy E_{p} stored in spring
Given: spring constant k = 100 N/m; compression distance x = 10 cm = 0.1 m Relationships: E_{p} = ½kx^{2} Solve: $${E}_{p}=\frac{1}{2}k{x}^{2}=\frac{1}{2}\left(100\text{\hspace{0.17em}}\text{N}/\text{m}\right)\times {\left(0.1\text{\hspace{0.17em}}\text{m}\right)}^{2}=0.5\text{\hspace{0.17em}}\text{J}$$Answer: The spring stores 0.5 J of potential energy.  Potential energy stored in a spring does not depend on how the spring was compressed. Therefore, in this new case, the stored energy is the same, 0.5 J.

Take a Quiz 